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PROBLEM #1 NCEES SPECIFICATION: DRINKING WATER DISTRIBUTION & TREATMENT TOPIC: UV DISINFECTION - BANKS OF LAMPS The manufacturer of a UV disinfection unit has provided the following constants for the RED equation: a = -0.80, b = -2.0, c = 0.18, d = 0.50, and e = 0.85. The following values apply to the water being treated: maximum UVA = 0.100 cm-1, minimum S/So = 0.8, maximum flow = 5 MGD. The validation factor for the model calibration with respect to Cryptosporidium is 1.8. How many banks of lamps will be required to disinfect this water to meet the requirements of the Long-Term 2 Enhanced Surface Water Treatment Rule?

A. 2 B. 4 C. 6 D. 8

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PROBLEM #1 ANSWER: A Expect a UV disinfection problem on your exam. This problem assumes that you’re familiar with RED and it’s associated formulae. Per the EPA, RED is "the UV dose derived by entering the log inactivation measured during full-scale reactor testing into the UV dose-response curve that was derived through collimated beam testing. RED values are always specific to the challenge microorganism used during experimental testing and the validation test conditions for full-scale reactor testing.” Because RED is based on the actual disinfection performance of the UV system, it is the most accurate and reliable manner in sizing UV systems. This problem also expects that you know that a 2-log (99%) inactivation of Cryptosporidium requires a dose of 5.8 mW/cm2. Here’s a handy UV dosage table:

UV Dose Values (mW/cm2) Log Inactivation

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

Cryptosporidium

1.6

2.5

3.9

5.8

8.5

12

15

22

Giardia

1.5

2.1

3.0

5.2

7.7

11

15

22

Virus

39

58

79

100

121

143

163

186

Step 1: Compute RED (reduction equivalent dosage).

RED = (Validation Factor)(Required Dosage) = (1.8)(5.8) = 10.44 mW/cm2

Step 2: Use RED to determine the number of lamp banks. c

S 1 R E D = (10 )(U VA ) (B e ) ( ( S0 ) f l o w ) a

d

b

c

S 1 R ED = (10 )(U VA ) (B e ) ( ( S0 ) f lo w ) a

b

d

→

10.44 = (10−0.80 )(0.100−2.0 )(0.80)0.18

1 (5)

0.50

(B 0.85)

→

Since we need a whole number to represent the amount of banks (B), we must round up to 2.

Therefore, the design will require two banks of UV lamps.

B = 1.65

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PROBLEM #2 NCEES SPECIFICATION: HYDROLOGY TOPIC: MULTI-OUTLET POND RISER Two rectangular orifices mounted in a concrete riser (shown below) are used to control flow from a stormwater detention facility. What is the flow (cfs) through the riser box if the water elevation in the pond is 104.5 feet? Assume orifices have 0.6 discharge coefficients and flow discharges without experiencing any submergence.

105.00 ft 1.0 ft W x 1.0 ft H 103.00 ft

1.20 ft W x 1.0 ft Basin bottom 2-ft dia. outlet culvert

A. 12.5 B. 14.7 C. 16.4 D. 18.9

100.00 ft

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PROBLEM #2 ANSWER: C This is a practical problem, and not hard to solve. Since the problem statement only asks for the flow at 104.5 ft pond elevation, you don’t need to worry about pond elevation decreasing over time, as outflow through the orifices occurs. This said, we need to solve for the flow out of each orifice, and add our results. Additionally, you can ignore the 2-ft diameter pipe in this problem, is it won’t impact the abovepond head analysis required to solve this problem.

Step 1: Use the submerged orifice equation for the top orifice.

Q = C A 2gh = (0.6)(1) (2)(32.2)(104.5 − 103.5) = 4.82c fs Note that h is measured from the center of the orifice opening.

Step 2: Use the submerged orifice equation for the bottom orifice.

Q = C A 2gh = (0.6)(1.2) (2)(32.2)(104.5 − 100.5) = 11.56c fs

Step 3: Add the flows together to solve the problem.

Answer is 4.82 cfs + 11.56 cfs = 16.38 cfs

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PROBLEM #3 NCEES SPECIFICATION: ANALYSIS & DESIGN TOPIC: FORCE ON A SPILLWAY CURVE Flow occurs over a spillway of constant section as shown below. What is the horizontal force (lb/ft) on the spillway per foot of spillway width (perpendicular to the spillway section)? Assume ideal flow.

A. 267 B. 300 C. 321 D. 365

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PROBLEM #3 ANSWER: A Force on a curve problems shows up regularly on the PM WRE portion of the exam, so you should expect to encounter it. This problem is tricky because the application is a spillway, and the problem is asking for a force per foot.

Step 1: Conduct an energy balance analysis between points 1 and 2.

P1 v12 P v2 + + z1 = 2 + 2 + z2 γ 2g γ 2g !

0! +

v12 v2 2 +4=0+ + 0.6 (2)(32.2) (2)(32.2)

Step 2: Use the principle of continuity to solve for v1 in terms of v2.

v1A1 = v2A2 Since we’re analyzing per foot of width, the area will be equal to the height of flow, in both cases, therefore

v1(4) = v2(0.6), and therefore, v1 = 0.15v2 ft/s Step 3: Back to the energy analysis equation of step 1.

0+

(0.15v2 )2 v2 2 +4=0+ + 0.6 (2)(32.2) (2)(32.2)

→

v2 = 14.97f t /s and v1 = 2.25 ft/s.

Step 4: Determine the horizontal force per foot width on spillway.

γh1(h1 /2) − γh 2(h 2 /2) − Fx = ρQ1(v1 − v2 ), where Q1 = whv1 = (1)(4)(2.25) (62.4)(4)(4/2) − (62.4)(0.6)(0.6/2) − Fx = (1.94)(1)(4)(2.25)(14.97 − 2.25) !

→

Fx = 267.4lb /f t

(r ight war d )

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PROBLEM #4 NCEES SPECIFICATION - WASTEWATER COLLECTION & TREATMENT TOPIC - INFILTRATION & INFLOW A city has the following sewer rates during the wet season: Day

Hour

Flowrate (MGD)

1

Midnight

3.7

4AM

2.5

8AM

7.1

Noon

6.8

4PM

6.3

8PM

7.0

Midnight

5.5

Yes

4AM

15.5

Yes

8AM

12.3

Yes

Noon

9.2

4PM

8.8

8PM

7.0

Midnight

4.5

4AM

4.9

8AM

8.3

Noon

8.6

4PM

9.3

8PM

9.1

Midnight

6.4

2

3

4

Storm Occurrence?

The dry weather flow averages 3.4 MGD. Excessive infiltration is defined as exceeding 4,500 gal/d-in-mi. If the collection system consists of 60 miles of 12-inch and 125 miles of 24-inch sewer, what is the total combined infiltration and inflow (MGD).

A. 10.1 B. 14.5 C. 15.2 D. 16.3

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PROBLEM #4 ANSWER: C Inflow and infiltration (I/I) is a topic you’re likely to experience on your exam. You must know the difference between the two, and how to solve for each, given data (table or chart).

Step 1: Calculate inflow.

Inflow is the difference between the peak flow caused by a storm, and the flow during the same period without a storm. This said, from the chart, the peak storm-related flow is on the second day at 4AM, at 15.5 MGD. Let’s compare that to the day before at the same time (4AM), and we get 2.5 MGD. We use the day before, as we can assume that a storm did not impact that flowrate; the following day may include some latency, which we don’t want to consider in this analysis. So, the inflow for this system is 15.5 MGD - 2.5 MDG = 13.0 MGD. Step 2: Calculate infiltration.

Infiltration is the difference between the average daily non-storm-influenced wet-weather and dry-weather flows. The problem statement provides the average for the dry season as 3.4 MGD. Now, we need to use the above wet season table to average a dry period (during the wet season) — we can use the first day again, as we know it isn’t storm-influenced. Averaging the first day’s flows, we get:

! avg = Q

3.7 + 2.5 + 7.1 + 6.8 + 6.3 + 7.0 = 5.6MGD 6

Therefore, the infiltration is 5.6 MGD - 3.4 MGD = 2.2 MGD. Adding, both inflow and infiltration, we get 13.0 MGD + 2.2 MGD = 15.2 MGD. *** BONUS MATERIAL *** What if the problem indicated that “excessive infiltration” is defined as exceeding 4,500 gal/d-inmi, and asked you to determine if this system’s infiltration is excessive? First, you need to compute the composite diameter-length of the collection system = (60 mi)(12-in)+(125 mi)(24-in) = 3,720 inch-miles. Next, we divide our wet-weather infiltration rate by the composite diameterlength = (2.2 x 106 gal)/(3,720 inch-mi) = 591.4 gal/day-in-mi. Since it is less than the 4,500 gal/ day-in-mi threshold, it is not considered “excessive.”

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PROBLEM #5 NCEES SPECIFICATION: HYDRAULICS - CLOSED CONDUIT TOPIC: BRANCHED RESERVOIR SYSTEM In the figure below, the valve F is partly closed, creating a head loss of 3.60 ft when the flow through the valve is 1 cfs. What is the length (feet) of the 10-inch pipe between reservoir A and junction B?

A. 645 B. 882 C. 1119 D. 1258

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PROBLEM #5 ANSWER: B Expect a branched (likely three) reservoir problem on the PE exam. Yes, some of these problems can be extremely difficult, but they can also be designed to be solved well-within 6 minutes. Note: in branched reservoir problems, velocity head can be considered negligible when compared to elevation and pressure heads, and therefore, can be considered equal to zero. Also, minor losses can be considered zero, unless otherwise stated in the problem.

Another note: we offer more branched reservoir problems in some of our other problem sets.

So if you want to master this topic, be sure to check them out.

Step 1: Use the information you have to begin. “C” values were given in the problem figure, and by considering the Hazen-Williams equation, we can solve for the energy gradient (s). Start by analyzing pipeline DB.

Q ! = 0.432CD 2.63s 0.54 = (0.432)(80)(1)0.63s 0.54 = 1c fs Therefore, s = 0.0014 ft/ft.

Step 2: Determine the head loss in pipeline DB (pipe friction + valve loss).

(h ! L )DB = (0.0014f t /f t)(1000f t) + 3.60f t = 5.01f t Step 3: Compute the elevation at junction B, then use that to determine the energy gradient and flow in pipeline BE. If ELE=0, the grade line elevation at B = 20 - 5.01 = 14.99ft. From there

s! BE = (14.99 − 0)/5000 = 0.0030 2.63 ! BE = (0.432)(120)(1) Q (0.003)0.54 = 2.25c fs

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PROBLEM #5 SOLUTION CONTINUED Step 4: Now we can solve QAB, sAB, and finally, LAB.

Q ! AB = QBE − QBD = 2.25 − 1 = 1.25c fs 1.25 = (0.432)(100)(10/12)2.63(sAB )0.54 ! ∴

sAB

=

0.0034

=

(hf )

AB

LAB

=

(20 − 15 − 2) , and LAB = 882 feet.

LAB

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PROBLEM #6 NCEES SPECIFICATION: HYDRAULICS - OPEN CHANNEL FLOW TOPIC - FLOWRATE THROUGH CULVERT The figure below shows the elevations of high water marks from a recent flood event. What was the discharge (cfs) through the culvert, if the culvert is a 54-inch reinforced concrete pipe with a Manning’s roughness coefficient of 0.013 and the inlet has a groove end, projecting (ke=0.02).

A. 190 B. 205 C. 220 D. 245

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PROBLEM #6 ANSWER: A A culvert problem is likely to appear on your exam. The math is simple, but there are many culvert scenarios (submerged vs. unsubmerged, full-flow vs. partial flow, inlet vs. outlet control) and equally as many nomographs that you can use to solve each type of scenario. The go-to design manual for culverts is “Hydraulic Design of Culverts”, third edition, by the U.S. Department Transportation Federal Highway Administration. It’s a large document, but covers both theory and technical details. Even more important, it provides you with the many nomographs you’ll need to solve these problems. This will be the first of many culvert problems we cover in our problem sets, so if you want more experience, be sure to check out our other problem sets. Here’s a link to the culvert design document: https://drive.google.com/file/d/ 1hKuuZzTZ9ujM4ee0bfD3lMkli4NT9aCn/view?usp=sharing Step 1: Ultimately, we’ll need to solve for the discharge rate twice, assuming both inlet and outlet control, then select the lower of the two flowrates. In both cases, we’ll need the value of HW, so we’ll start there.

H ! W = 286.88 − (250)(0.01) − 275.83 = 10.8f t Step 2: Let’s first assume the inlet controls the flowrate. This said, we can go to the “Chart 1B” nomograph (see next page), as it specifically was designed for circular concrete pipe inlet control, and we will solve for Q once we know D and HW/D. Before using the nomograph, let’s solve for HW/D.

HW 10.8 = = 2.4 D 4.5 !

Step 3: Use the “Chart 1B” nomograph to solve for Q. See the image on the next page. Enter the chart from the right where the third column is equal to 2.4. Use the third column, as the chart indicates that the entrance type for that column is “groove and projecting” — just like in our problem statement. Continue to draw the line to the left to the first column, then connect the line to the pipe diameter. Where it intercepts the Q-value column is your answer for flowrate. So, in this example, the flowrate is approximately 245 cfs.

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PROBLEM #6 SOLUTION CONTINUED

Step 4: Next, determine the discharge assuming outlet control. In order to solve for Q, first compute the tailwater depth and the head loss.

T ! W = 282.3 − 275.82 = 6.47f t H ! = H W − T W + S0 L = 10.8 − 6.47 = (250)(0.001) = 4.58f t

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PROBLEM #6 SOLUTION CONTINUED Step 5: Now we can solve for velocity of flow, which we’ll convert into a flowrate.

29n 2 L v 2 (29)(0.013)2(250) H = 1 + 0.02 + ! = 1 + Ke + 1.33 R 1.33 ] 2g [ 4.5 ( 4 )

v2 = 4.58 (2)(32.2)

Solving for v, we get 11.94 ft/s, which we can easily convert into a flowrate.

π (4.5)2 ! = (11.94) Q = 190c fs ( 4 ) Step 6: Determine the correct flowrate — inlet our outlet control.

Use the lower value of the inlet and outlet control flowrates for the same headwater depth; therefore, the flowrate through the culvert is 190 cfs, since 190 < 245.

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PROBLEM #7 NCEES SPECIFICATION: GROUNDWATER & WELLS TOPIC: UNCONFINED AQUIFER - GROUNDWATER FLOW CHARACTERISTICS BETWEEN 2 CHANNELS A canal was constructed running parallel to a river 1500 ft away. Both fully penetrate an unconfined sand aquifer with a hydraulic conductivity of 1.2 feet per day. The area is subject to rainfall of 1.8 feet per year and evaporation of 1.3 feet per year. The elevations of water in the river and canal are 31 ft and 27 ft, respectively. What is the maximum water table elevation (ft)?

A. 25 B. 28 C. 35 D. 39

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PROBLEM #7 ANSWER: D This is a classic groundwater problem, and something very similar to it can show up on your exam. It’s often helpful to visualize groundwater questions. In this case, the following image depicts the scenario.

In this image, d represents the distance to the “water divide” (the inflection point of the water table height curve, and location of the maximum height. W represents the net addition or loss of water through infiltration, evaporation, and evapotranspiration. This said, the problem statement is asking us to solve for d first, then hmax.

Step 1: Solve for W. W = 1.8 ft/yr infiltration - 1.3 ft/yr evaporation = 0.5 ft/yr = 0.0014 ft/d accretion.

Step 2: Solve for d (for unconfined aquifer between 2 channels).

d=

L K(h12 − h 2 2 ) 1500 (1.2)(312 − 272 ) − = − = 684f t from the river.

2 w 2L 2 (0.0.0014)(2)(1500)

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PROBLEM #7 SOLUTION CONTINUED Step 3: Solve for the corresponding hmax.

hma x =

(h12 − h 2 2 )x w h1 − + (L − x)x = L K 2

312 −

(312 − 272 )(684) 0.0014 + (1500 − 684)684 = 38.8f t 1500 1.2

Note that this elevation is greater than either the river or channel, which is expected and represented in the illustration above.

*** BONUS MATERIAL ***

Daily flow into River (x = 0):

Q ! width =

K(h12 − h 2 2 ) L (1.2)(312 − 272 ) 1500 −w −x = − (0.0014) − 0 = − 0.957f t 3 /d /f t ( ) ( ) 2L 2 (2)(1500) 2

The negative value indicates it’s flowing from right to left, into the river.

Daily flow into canal (x = 1500):

K(h12 − h 2 2 ) L (1.2)(312 − 272 ) 1500 Q −w −x = − (0.0014) − 1500 = 1.143f t 3 /d /f t ! width = ( ) ( ) 2L 2 (2)(1500) 2 The positive value indicates it’s flowing from left to right, into the canal.

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PROBLEM #8 NCEES SPECIFICATION: ANALYSIS & DESIGN TOPIC: SPILLWAY DISCHARGE WITH COEFFICIENT CORRECTION A spillway has been designed for a head of 2.80 m with a length of 200 m and a discharge coefficient equal to 0.49. What is the maximum discharge that can be passed over this spillway without cavitation, assuming Hmax = 1.65HD? Use the figure below, as necessary.

A. 3,108 B. 4,663 C. 5,741 D. 6,009

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PROBLEM #8 ANSWER: A Less-than-simple spillway problems appear on the exam regularly. Many of the problems require using discharge coefficient and coefficient correction factor charts to solve the problem.

These charts will be provided during the exam. We offer many example problems both involving and not involving charts (see other problem sets for more spillway problems). This is a chartdependent problem. Note: when a weir discharges at a head different from the design head, the flow differs from ideal, and the discharge coefficient changes â&#x20AC;&#x201D; therefore, it must be corrected.

Step 1: Solve for Hmax then use the given chart to determine the corrected C-value for the new maximum head. ! ma x = 1.65HD = (1.65)(2.80) = 4.62m (greater than the design head, so correction needed) H

For !

Hma x 4.62 = = 1.65 HD 2.8

â&#x2020;&#x2019;

C = 1.08 CD

From the given chart we can assume the curve approaches 1.08 once the head ratio reaches 1.65. ! corrected = (CD )(cor r ect ion C

fa ctor) = (0.49)(1.08) = 0.53

Step 2: Solve for discharge over the crest.

Q ! ma x =

2 2 C 2gL H 3/2 = (0.53)( (2)(9.81)(200)(4.62)3/2 = 3,108m 3 /s (3) (3)

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PROBLEM #9 NCEES SPECIFICATION: DRINKING WATER DISTRIBUTION & TREATMENT TOPIC: HARDNESS & SOFTENING SPLIT SYSTEM DESIGN Determine the chemical dosages (lime and soda, respectively, as CaCO3) for a split treatment system to soften the following water. The finished water shall have maximum magnesium hardness of 40 mg/L as CaCO3 and a total hardness of 80 mg/L as CaCO3. Assume excess lime is not necessary.

Constituent

mg/L as CaCO3

CO2

20

Ca2+

240

Mg2+

60

HCO3-

198

Cl-

75

SO42-

75

Na+

55

A. 240, 39 B. 282. 75 C. 282, 39 D. 240, 75

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PROBLEM #9 ANSWER: A All softening problems require strict adherence to a process, and split problems aren’t any different.

Will all softening problems, we turn to Davis and Cornwell’s Introduction to Environmental Engineering book for the process. First let’s imagine the split system at-hand.

A portion of the flow is treated, another is bypassed, and then they are both mixed to meet the final hardness requirements. Next, let’s take a look at the Davis and Cornwell process:

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PROBLEM #9 SOLUTION CONTINUED These aforementioned dosage schemes are for when Mg2+ concentration is less than 40 mg/L as CaCO3 and no split treatment is required. So, these dosage schemes are only appropriate when the question asks you to solve for lime and soda dosages for softening water to a practical solubility limit. These donâ&#x20AC;&#x2122;t work for our problem statement, since the Mg2+ concentration is not less than 40 mg/ L as CaCO3, and we do need to design a split system. However, keep the aforementioned dosage schemes handy for your exam. Now, the dosing schemes we need are

These dosage schemes are for when the Mg2+ concentration is greater than or equal to 40 mg/L as CaCO3, and a split treatment design is required. Note that these schemes require softening to the practical limits in the first stage of the split-flow scheme.

So, let us use these dosing schemes to proceed with the problem.

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PROBLEM #9 SOLUTION CONTINUED Step 1: Draw the bar chart (only necessary constituents).

20

0

235 Ca2+

295 Mg2+

CO2 HCO3-

Cl198

273

Step 2: Solve for TH, CH, and NCH. TH = 295 CH = 273 NCH = 295 - 273 = 22 mg/L as CaCO3 Step 3: Solve for lime and soda for the first stage, to the theoretical solubility limits.

We now need to compare are bar chart to those in the second set of dosing schemes. Our bar chart looks most like case (a) where Ca2+ overshadows HCO3-, so weâ&#x20AC;&#x2122;ll use that process for this portion of the flow.

1. add lime equal to CO2 = 20 2. add lime equal to HCO3- = 198 3. add lime equal to Mg2+ = 60 4. add soda equal to (Ca2+ + Mg2+) - HCO3- = (235 + 60) - 198 = 97 mg/L as CaCO3.

5. add excess lime = 0 (problem statement indicated excess lime is not necessary) Total lime: 278 as CaOH3 Total soda: 97 as CaOH3 It is important to note that this process provides a finished water with a Mg2+ concentration of 10 mg/L as CaCO3. This will be necessary to know in the next step.

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PROBLEM #9 SOLUTION CONTINUED Step 4: Calculate the split

X=

Mgf − Mgi Mgr − Mgi

Where Mgf = final (post-mix) Mg concentration mg/L as CaCO3 (40 mg/L as CaCO3, unless specified otherwise) Mgi = Mg concentration from first stage mg/L as CaCO3 (10 mg/L as CaCO3, unless specified otherwise) Mgr = raw water Mg conc. mg/L as CaCO3 (this was provided in the problem statement table)

X=

Mgf − Mgi Mgr − Mgi

=

40 − 10 = 0.60 60 − 10

Looking back at our schematic, this means that 60 percent of the flow is bypassed with 40 percent of the flow being treated in the first stage.

Step 5: Determine the final overall hardness.

Raw Water Hardness + Treated Water Hardness = (0.6)(295) + (0.4)(40) = 193 mg/L as CaCO3. This mixed-water total hardness is higher than the 80 mg/L as CaCO3 specified in the problem statement.

Since the Mg2+ hardness is already accurate (40 mg/L as CaCO3), we’ll have to reduce the Ca2+ hardness down to 40 mg/L as CaCO3 — this will give us a total of 80 (40 + 40) mg/L as CaCO3.

Step 6: Reduce the Ca2+ hardness in the mixed water.

How much calcium will we need to remove to be left with 40 mg/L as CaCO3? Well, (235 - x) = 40, or x = 195. This level of removal can be done with removing most of the bicarbonate (HCO3-). This said, we’ll need to add more lime: 1. add lime to remove CO2 (to raise pH): 20 2. add lime to remove HCO3-: 195 So, an additional 20 + 195 = 215 mg/L as CaCO3 is added to the mixed water. No soda needed.

SQRPGZ PROBLEM #9 SOLUTION CONTINUED Step 7: Summarize total chemical additions in proportion.

Lime: (0.4)(278) + (0.6)(215) = 240.2 mg/L as CaCO3 Soda: (0.4)(97) + (0.6)(0) = 38.8 mg/L as CaCO3

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PROBLEM #10 NCEES SPECIFICATION: HYDRAULICS - CLOSED CONDUIT TOPIC: PIPE NETWORK HARDY CROSS The pipe network below shows known flowrates and flow direction for all flows in and out of the network, along with estimated flowrates and flow direction for all flows within pipes 1 through 7.

Using the figure (below) and table (next page), and applying the Hardy Cross Method, what is the estimated flowrate (cfs) in pipe 2 after one iteration?

Q = 14cfs

Q = 6.64cfs

Pipe 1

L = 2000ft, D = 18in, Q = 7cfs

Pipe 4

L = 1100ft

D = 24in

Q = 7cfs

Q = 4.81cfs

A. 3.50 B. 4.73 C. 5.18 D. 6.08

Loop 1

Pipe 2

L = 900ft

D = 24in

Q = 3.5cfs

Pipe 3

L = 2800ft, D = 12in, Q = 2.19cfs

Pipe 5

L = 2200ft, D = 12in, Q = 3.5cfs

Loop 2

Pipe 6

L = 750ft

D = 18in

Q = 3.14cfs

Pipe 7

L = 2600ft, D = 24in, Q = 5.69cfs

Q = 2.55cfs

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PROBLEM #10 ANSWER: C Whether quantitative or qualitative, you need to be prepared for a pipe network problem on your exam. This happens to be quantitative, and although it’s a Hardy Cross analysis problem, it is not requiring you to perform multiple iterations to solve the proposed problem. Rather, you’ll need to only perform one single iteration. This said, the key to the Hardy Cross Method is establishing and maintaining a process. In answering this problem, we’ll share our process. It’s straight-forward, and we’re sure it will help you through the exam. Note: we offer many more pipe network problems in our other problem sets. We encourage you to familiarize yourself with those problems so you can be versatile in solving these type of problems.

Step 1: Create tables for each loop that generate information used to compute a estimated flow correction.

Before we construct our tables, we need to first assume a clockwise or counterclockwise direction in each loop. We’ll assume clockwise. This means that all flows within the loop that are clockwise will be expressed in the table with a positive flow value, whereas those that flow opposite the clockwise direction will be expressed in our tables with a negative flow value (see column 4 in our tables). Note: pipe 2 is part of both loops, and flows clockwise (positive flow value) in loop 1, and flows counterclockwise (negative flow value) in loop 2.

Loop 1 (1) Pipe

(2) L (ft)

(3) D (ft)

(4) Q (cfs)

(5) Unit hf (ft/ft)

(6) hf (ft)

(7) hf/Q (s/ft2)

Given

Given

Given

Given

From pipe diagram

(2) x (5)

(6) / (4)

1

2000

18

7.00

0.0034

6.80

0.971

2

900

24

3.50

0.000235

0.212

0.061

3

2800

12

-2.19

-0.0029

-8.120

3.708

4

1100

24

-7.00

-0.00085

-0.935

0.134

Sum = -2.043

Sum = 4.874

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PROBLEM #10 SOLUTION CONTINUED Loop 2 (1) Pipe

(2) L (ft)

(3) D (ft)

(4) Q (cfs)

(5) Unit hf (ft/ft)

(6) hf (ft)

(7) hf/Q (s/ft2)

Given

Given

Given

Given

From pipe diagram

(2) x (5)

(6) / (4)

5

2200

12

3.50

0.007

15.400

4.400

6

750

18

-3.14

-0.00078

-0.585

0.186

7

2600

24

-5.69

-0.00057

-1.482

0.260

2

900

24

-3.50

-0.00023

-0.207

0.059

Sum = 13.126 Sum = 4.905

Note: In this problem you were provided with a unit head loss chart to ultimately solve for the headloss in each pipe; however, if this table isn’t provided, you may be provided with DarcyWeisbach friction factors or Hazen-Willams “C” values. In any case, you will need to solve for the headloss. It’s also important to note that the positive or negative will carry over to hf from Q, in the tables.

Step 2: Solve for the flow correction factor in each loop.

! ΔQ =−

Σhf [

nΣ

( Q )]

(n=1.85 for Hazen-Williams, and n=2 for Darcy-Weisbach)

hf

! ΔQ loop1 = −

−2.043 = 0.23c fs (1.85)(4.874)

ΔQ ! loop2 = −

13.126 = − 1.45c fs (1.85)(4.905)

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PROBLEM #10 SOLUTION CONTINUED Step 3: Apply the correction factors to pipe 2 to solve the problem.

The next step in the Hardy Cross Method is to apply the flow correction factors to the initial estimates. In the case of pipe 2, since it is part of both loops, we need to apply both correction factors to it. From the loop 1 perspective, 3.50cfs + 0.23cfs - (-1.45cfs) = 5.18cfs. Since the loop 1 correction factor is positive, it backs-up our assumption that flow is moving clockwise.

This said, weâ&#x20AC;&#x2122;ve added it to the initial flow estimate. Since the loop 2 assumption of clockwise is actually counter clockwise from the loop 1 perspective, we subtract the loop 2 correction factor from our initial flow estimate.

If we apply both correction factors to pipe two from the loop 2 perspective, the flow should be equal in magnitude and opposite in direction. This said, we get -3.50cfs - (+0.23cfs) + (-1.45cfs) = -5.18cfs. The answer checks. Therefore, 5.18 cfs is the next flow estimate.

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PROBLEM #11 NCEES SPECIFICATION: WATER QUALITY TOPIC: CRITICAL OXYGEN DEFICIT (WITH ALGAL & BENTHIC OXYGEN DEMANDS) An outfall discharges wastewater into a slow-moving river that the has a mean velocity of 3 cm/s and average depth of 3 m. After initial mixing, the DO concentration of the river is 9.5 mg/L, the saturation of oxygen is 10.1 mg/L, the ultimate BOD of the mixed river water is 20 mg/L, the reaeration rate constant is 0.48 day-1, and the deaeration rate constant is 0.72 day-1. During the night, algal respiration exerts an oxygen demand of 2 g/m2-day and sludge deposits downstream of the outfall exert a benthic oxygen demand of 4 g/m2-day. What is the critical oxygen deficit of the river?

A. 7.2 B. 8.1 C. 8.6 D. 9.2

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PROBLEM #11 ANSWER: B This is an oxygen dynamics problem. It’s tricky because the problem statement introduced the volumetric oxygen demand rates for algal respiration Sr and benthic consumption Sb. This said, traditional equations for critical distance and critical oxygen demand need to be revised.

Step 1: Solve for the distance to the critical oxygen sag point.

x! c =

k D (k − kd ) + (Sr + Sb )(kr − kd ) k v ln r − r a r kr − kd [ kd ] kd 2 La

Before we can plug and chug, we must find the values for each variable. kr = 0.48 day-1 kd = 0.72 day-1 Da = 10.1 mg/L - 9.5 mg/L = 0.60 mg/L (initial oxygen deficit after mixing) La = 20 mg/L (initial ultimate BOD after mixing) Sr = -(2 g/m2-day)/(3 m) = -0.667 g/m3-day Sb = -(4 g/m2-day)/(3 m) = -1.33 g/m3-day v = 3 cm/s = 2,592 m/d Therefore,

x! c =

2592 0.72 (0.72)(0.60)(0.72 − 0.48) + (−0.667 − 1.33)(0.72 − 0.48) ln − = 4,951m 0.72 − 0.48 [ 0.48 ] (0.48)2(20)

Step 2: Solve for the critical oxygen deficit at distance xc.

D ! c=

k x S + Sb kr 0.48 (0.48)(4951) −0.667 − 1.33 La ex p − d c − r = (20)ex p − − = 8.1mg /L ( V ) ( ) kd kr 0.72 2592 0.72

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PROBLEM #12 NCEES SPECIFICATION: HYDROLOGY TOPIC: UNCONNECTED IMPERVIOUS SURFACE RUNOFF A 3-acre development site is comprised of 1 acre of impervious surface (NRCS CN = 98) and 2 acres of lawn and woods (NRCS CN = 65). Runoff from the entire impervious surface sheet flows onto the pervious portion of the site before entering the siteâ&#x20AC;&#x2122;s drainage system. What is the total runoff volume (cf) to the drainage system for a 1.25-inch runoff event.

A. 274 B. 444 C. 521 D. 581

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PROBLEM #12 ANSWER: D There are mainly two NRCS-based ways to solve runoff problems that involve unconnected impervious surface areas that are upstream of pervious area: (1) NRCS TR-55 Methodology (see manual), which are applicable only to sites with less than 30 percent total impervious coverage and the downstream pervious area must be at least twice as large as the unconnected impervious area; and, (2) the twostep technique, which involves using the NRCS runoff equation, but treats resultant runoff from the upstream impervious unconnected area as additional rainfall on the downstream pervious area.

Since the impervious area is 33 percent of the total area, and greater than 30 percent, we must use the two-step technique. Let’s look at a picture of the circumstances.

Step 1: Compute the runoff volume from the impervious area.

Impervious area S =

1000 1000 − 10 = − 10 = 0.20in CN 98

(P − 0.2S )2 (1.25 − (0.2)(0.2))2 Impervious area runoff = Q = = = 1.04in P + 0.8S 1.25 + (0.8)(0.2)

Runoff volume = (1.04in)

1f t 43,560f t 2 (1a cr e) = 3,775f t 3 ( 12in ) ( 1a cr e )

Equivalent rainfall on downstream pervious area:

3,775f t 3 1a cr e = 0.043f t = 0.52in ( 2a cr es ) ( 43,560f t 2 )

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PROBLEM #12 SOLUTION CONTINUED Step 2: Compute the total runoff volume from the pervious area, including runoff from impervious upstream area.

Total effective rainfall = direct rainfall + unconnected impervious area runoff = 1.25 in + 0.52 in = 1.77 in

Pervious area S =

1000 1000 − 10 = − 10 = 5.38in CN 65

(P − 0.2S )2 (1.77 − (0.2)(5.38))2 = = 0.08in Pervious area runoff = Q = P + 0.8S 1.77 + (0.8)(5.38) 1f t 43,560f t 2 (2a cr es) = 581f t 3 Pervious area runoff volume = (0.08in) ( 12in ) ( 1a cr e )

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PROBLEM #13 NCEES SPECIFICATION: HYDRAULICS - OPEN CHANNEL TOPIC: COMBINED EFFECTS OF CHANNEL CONSTRICTION AND DEPRESSION ON WATER SURFACE ELEVATION If 1.4 m3/s of water flows uniformly in a channel of width 1.8 m at a depth of 0.75 m, what is the change in water surface elevation (m) at a section that is contracted to 1.2 m width and simultaneously experiences a 6-cm depression in the bottom of the channel?

A. 0.073 B. 0.377 C. 0.737 D. 0.808

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PROBLEM #13 ANSWER: A This problem is unique. Normally, open channel problems add a hump to the channel bottom or constrict the channel width, and ask that you perform energy balance analyses; however, this problem combines width constriction with channel bottom adjustment. Furthermore, the channel bottom adjustment isn’t the addition of a hump, rather, a depression. This is the level of difficulty you should expect in the PM portion of the WRE exam.

Step 1: Conduct an energy balance analysis at points 1 (upstream) and 2 (point of constriction and depression).

y1 +

v12 v2 + z1 = y2 + 2 + (−z2 ) 2g 2g

→

E1 + z1 = E2 − z2

If we assume that our datum is the upstream channel bottom, then z1 = 0, and therefore we get

E1 + z2 = E2 we already know the value for z2 (depression value), and we can solve for E1, leaving us with solving for E2 as a function of y2. Afterwards, we can plug everything into the above equation to solve for y2, then use that to determine the drop in surface water elevation.

Step 2: Solve for E1, where the channel width is 1.8 m.

E1 = y1 +

v12 = 0.75m + 2g

1.4m 3 /s [ (0.75m)(1.8m) ]

(2)(9.81)

2

= 0.8048m

Step 3: Solve for E2 as a function of y2. Remember the channel width at point 2 has been constricted to 1.2 m.

E2 = y2 +

1.4 [ ( y2 )(1.2) ]

2

v2 0.0694 = y2 + = y2 + 2g (2)(9.81) y2 2 2

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PROBLEM #13 SOLUTION CONTINUED Step 4: Substitute the results from steps 2 and 3 into the equation established in step 1. Make sure the 6-cm depression is converted to m.

0.8048 +

6 0.0694 = y2 + 100 y2 2

There are two roots to this equation, y2 = 0.737 and y2 = 0.377. In order to figure out which one is the correct answer, we can compare each to the critical depth at point 2.

Step 5: Determine the critical depth at point 2.

yc =

2 Ec = 3

3

2

Q = g w2

3

2

q = g

1.4 3 ( 1.2 )

9.8

2

= 0.517m

(where q is the flow per unit width, w)

If we compare our roots from step 4 to the critical depth, we see that one is less (supercritical) and the other is greater (subcritical). Since this is a channel depression (rather than a hump), we know that the depth cannot be less than the critical depth. Therefore, the depth of flow at point 2 is 0.737m, and the change in water surface elevation is 0.75 - (0.737 - 6/100) = 0.073 m (drop).

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PROBLEM #14 NCEES SPECIFICATION: WASTEWATER COLLECTION & TREATMENT TOPIC: LENGTH OF COMBINED SEWER OVERFLOW SIDE WEIR What is the length of a weir (m) that must be placed in an existing 1200-mm pipe, which will be used as a combined sewer, if the maximum wet-weather flow is 1.9 m3/s and maximum allowable wet-weather flow to the treatment plant is not to exceed 0.7 m3/s. The pipe slope is 0.003, the Manningâ&#x20AC;&#x2122;s n is 0.013, and the maximum dry-weather flow is 0.14 m3/s. Assume a ratio of incoming to outgoing head at the weir to be 20. Use the charts below, as necessary.

A. 5.5 B. 12.5 C. 15.7 D. 18.2

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PROBLEM #14 ANSWER: B During wet-weather events, combined sewers may need to overflow some volume, in order to deliver a flowrate to the treatment plant that is equal to or under treatment capacity. A side weir is a weir that is parallel to the wastewater flow, and located in the side of the sewer pipe. The weir should be high-enough to prevent dry-weather discharge. Flow over the weir depends primarily on the depth of flow above the weir in the adjacent channel. The rate of discharge over the weir varies along the crest because of the change in depth resulting from the diversion of water without an appreciable loss of energy.

This problem will require you to analyze dry- and wet-weather flow conditions, using a hydraulic elements chart. From there, youâ&#x20AC;&#x2122;ll use side-weir specific formulae to answer this question. The math is easy, the process is tedious â&#x20AC;&#x201D; this is a great example of PM WRE problem.

Step 1: Compute maximum capacity of 1200-mm pipeline.

! full = Q

0.312 8/3 1/2 0.312 D S = (1.2)8/30.0031/2 = 2.1m 3s n 0.013

The corresponding velocity (flowing full) is 1.89 m/s.

Step 2: Compute the flow characteristics at 1.9 m3/s (given in problem statement).

Depth of flow: Q/Qfull = 1.9/2.1 = 0.905. The corresponding d/D value is 0.73, therefore the depth of flow dn = (0.73)(1200 mm) = 876 mm = 0.876 m.

Velocity: When d/D = 0.73, V/Vfull = 1.14. Therefore, Vn = (1.14)(1.89) = 2.15 m/s.

Step 3: Compute the flow characteristics at 0.14 m3/s (given in problem statement).

Depth of flow: Q/Qfull = 0.14/2.1 = 0.07. The corresponding d/D value is 0.20, therefore the depth of flow dn = (0.20)(1200 mm) = 240 mm = 0.240. Since all of the dry weather flow must be retained, the weir height must be at least 0.240 m above the bottom of the pipe.

Velocity: When d/D = 0.20, V/Vfull = 0.56. Therefore, Vn = (0.56)(1.89) = 1.06 m/s.

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PROBLEM #14 SOLUTION CONTINUED Step 4: Compute the specific energy at the upstream end of the weir, with the Ackers’ equation.

E ! w=

1.2vn 2 + (dn − c) 2g

where: Ew = specific energy of flow relative to crest of weir (m,ft) Vn = normal velocity in approach channel during wet-weather event (m/s, ft/s) dn = normal depth in approach channel during wet-weather event (m, ft) c = height of weir above channel invert (m, ft) Therefore,

E ! w=

1.2vn 2 (1.2)(2.15)2 + (dn − c) = + (0.876 − 0.240) = 0.92m 2g (2)(9.81)

Step 5: Compute the required weir length with the chart given in the problem statement.

The problems statement gave us a n2 value of 20, therefore, if we solve for c/Ew, we can then use the chart to solve for L/B, which will get us to the length (L). c/Ew = 0.24/0.92 = 0.26. Using the chart, L/B is approximately equal to 15. Therefore, 15 = L/ (1.2 m), or L = 12.5 m.

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PROBLEM #15 NCEES SPECIFICATION: HYDRAULICS - CLOSED CONDUIT TOPIC: PUMP PLACEMENT TO AVOID CAVITATION (WITH SAFETY FACTOR) A pump supplies 10 cfs of water at 90oF from a pit with a constant water elevation experiencing 12 psi of atmospheric pressure. Suction friction loss is 4.0 ft at the operating point. The net positive suction head required (NPSHR) is 11.1 ft. What is the maximum height (feet) the pump centerline can be installed above the pit water level to provide a safety factor of 1.25 against cavitation?

A. 6.5 B. 7.3 C. 8.2 D. 8.9

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PROBLEM #15 ANSWER: C There are many types of pump questions that can make it to the exam (note: we cover each type of pump question in our many problem sets). This is a classic NPSHA-HPSHR problem of avoiding cavitation through pump placement. The major trick in this question is the safety factor. We will use the following equation to solve this problem:

NPSH A !

=

hatm + hz(s) − hf (s) − h vp

Step 1: Solve for the atmospheric head.

h! atm =

(12psi )(12in /f t)2 62.4lb /f t 3

= 27.7f t

Step 2: Solve for the pressure head, based on the vapor pressure of water at 90-degrees F. Hint, you’ll need to have a table to determine the vapor pressure of 100.7 psf.

h! vp =

100.7ps f

62.4lb /f t 3

= 1.61f t

Step 3: Determine the NPSHA. NPSHA = (safety factor)(NPSHR) = (1.25)(11.1ft) = 13.88 ft

Step 4: solve for the elevation difference.

h! z(s) = NPSH A − hatm + hf (s) + h vp = 13.86 − 27.7 + 4.0 + 1.61 = − 8.21f t Therefore, the pump centerline must be no more that 8.2 feet above the water level in the pit.